Alright, I'm new to this "electricity" thing, and I need some help.I'm trying to make a small, personal laser harp. I got eight laser pointers from WalMart that I would like to use as lasers. I have the Arduino program that takes the resistance in different photoresistors to see if there is a laser being pointed at them, and plays a note depending on the photoresistor inputs.The lasers themselves take 3 LR44 batteries (Small 1.5 V batteries) to function. If I have 8 in series, what would be the best way to power them? Can they be powered by the Arduino through creative use of resistors and/or anything else? Or should I find a way to integrate their original batteries into the loop? (I have TONS of extra LR44 batteries...)Also- is there any way to have my speaker play more than one tone at the same time, or at the very least to simulate the playing of 2 notes?Any comments, questions, concerns, ideas, or help that you all can think of? I appreciate any help or constructive criticism, and I can't wait until I can get enough to help others on the board! Thank you!

The 3 batteries will be in series, giving a total of 3*1.5 = 4.5V. This is below the voltage supplied by the Arduino's 5V pin, so that could feasibly be used to supply the power.

However, first you will need to know the current draw of the lasers. I would recommend removing the batteries and connecting them somehow with wires so you can connect an ammeter in series with them to measure the current draw when the laser is on. Only then will you be able to determine if the Arduino is capable of providing the power.

If the total current draw for ALL the lasers together is greater than a hundred milliamps or so then I would forget it - the voltage regulator on the Arduino, while rated at 800mA, would struggle to dissipate the heat and will get hot.

In this case an external power supply will be required (either the existing batteries, or a wall wart that can provide 4.5V).

If the current is low enough, you can use Ohm's law to calculate the resistance *per laser* to use to drop the 0.5v from the 5v supply to 4.5v.

Ok, thank you for all of your help so far!I tested the draw with my multimeter, on the 200mA setting, and I got around 14 or 15. What exactly does this mean for the draw? That there is 200*15 = 3000 mA draw, or what?And if I were to use an outide source of power, how would I be able to get the arduino to tell it when to release power? For instance, if I used a 9V battery to power the lasers, how would I, using the arduino, be able to tell it to turn on or off?

15 on the 200mA range means 15mA. The 200mA range can measure between 0 and 200mA.

So, you have 15mA per laser - multiply that by the number of lasers, and you have the total current.

But at 15mA for a laser you are inside the ratings for an Arduino IO pin. You just need to drop 0.5V at 15mA - which is R=V/I = 0.5/0.015 = 33? resistor. It looks to me like you can treat it just like an LED.

However, I am no expert on lasers, so I for one would feel happier if someone else could corroborate this before you go and blow up your Arduino

And if that's the case, and I connect them them all in series with the Arduino ground at the end, they'll all just pull what they need and it'll just work all fine and dandy?Also, does it matter where in the circuit I place the resistor?

Our assets consist of my Arduino Uno, 8 laser pointers (In varying stages of being taken apart), varying photoresistors, and a breadboard to put them all on (I'll move onto an actual board when I'm done with the planning). I would power the Arduino with a 9-volt battery, and I don't believe I have any regulators or transistors at my house. I could pick some up at Radioshack though...

Powering everything with a 9V battery will mean you'll be going through batteries every couple hours. A 9V is ~600mah, meaning it can provide 600ma for one hour, or 300 for two, etc. You'd be best off locating a 9V power supply and plugging it into the wall, but I can understand that if you're on the road with the harp that it would be handier powering it from a 9V.

With a 9V supply you can put a pair of lasers in series and four pairs in parallel.

It might also be worth a test to see if the lasers illuminate well enough at 3V; 2 AA batteries (~2000mah) would give you >8 hours powering all eight lasers and would be a lot cheaper than 9Vs. I suggest 3V because I don't think the 'shack stocks 3xAA holders.

Considering this is a first time, mostly "just because" project, a couple hours should be fine for me, and it's nice and simple because I have the holders, and I know it'll work perfectly. I don't expect this thing to be on more than about 10 minutes at a time.How would I go about finding a 9-volt power supply that plugs into a wall? Are they called something particular?Also, why a pair in series and four pairs in parallel? I would only want 8 actually, but why that configuration, instead the 8 in series as suggested earlier?

Considering this is a first time, mostly "just because" project, a couple hours should be fine for me, and it's nice and simple because I have the holders, and I know it'll work perfectly. I don't expect this thing to be on more than about 10 minutes at a time.How would I go about finding a 9-volt power supply that plugs into a wall? Are they called something particular?Also, why a pair in series and four pairs in parallel? I would only want 8 actually, but why that configuration, instead the 8 in series as suggested earlier?

Also, why a pair in series and four pairs in parallel? I would only want 8 actually, but why that configuration, instead the 8 in series as suggested earlier?

8 in series required high voltage : V = N x 3 V = 24 V, or even more. I'd suggest you connect all of them in parallel, and run from on-board +5V with small resistors in series with each laser to drop 0.5V as it was suggested. You wouldn't be able to get stable brightness connecting two units in series, though it may be o'k with switching power supply (price ^). With battery running down, or cheap wall-wart power supplies you would get a lot of troubles to keep current more or less constant.